How do you find the limit of $(x-3)/(x^2+4)$ as $x->1$ ?

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Answer 1 · 2016-10-27T21:26:29

$lim_{x rarr 1} {x-3}/{x^2 + 4} = - 2/5$

When we have not any indeterminate form and limit exists, the value of a limit $lim_{x rarr a} f(x)$ it's just $f (a)$ . It means that:

$lim_{x rarr 1} {x - 3}/{x^2 + 4} = {1 - 3}/{1^2 + 4} = - 2/5$ .

How do you find the limit of (x-3)/(x^2+4)(x-3)/(x^2+4) as x->1x->1?